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Proof step 31 from proof maxinf-skolemf (previous step)

np(n1), n < n1, np(n0), ∀ n1, n2, np. n1 - n2 < n - n0 → np(n2) → (∃ n. np(n) ∧ (∀ n0. n < n0 → ¬ np(n0))) ∨ (∃ n0. n1 < n0 ∧ np(n0)) ⊦ n1 = 0, ∃ n. np(n) ∧ (∀ n0. n < n0 → ¬ np(n0)), ∃ n0. n < n0 ∧ np(n0)

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 3 terms: n1

Additional information

Used terms: n1